I have a one dimensional quantum mechanical system composed by a particle in a potential $$V(x) = -\frac{1}{ma^2} \textrm{sech}^2(x/a)$$ The problem gives me the eigenstates $$\psi_k(x) = \frac{ika -\tanh(x/a)}{ika+1} \frac{e^{ikx}}{2 \pi}$$ so that the eigenvalues are $E_k = k^2/2m$. Now I have to compute the integral $$\int_{-\infty}^{\infty} dk \langle x| \psi_k \rangle \langle y|\psi_k\rangle$$ and the problem asks me if the $\psi_k$ with $k \in \mathbb{R}$ form a basis of the Hilbert space.

I have tried to compute the integral with the residue theorem. The integral splits in four parts, one of these part is $$\int_{-\infty}^{\infty} \frac{k^2 a^2}{k^2a^2+1} e^{ik(x-y)} dk$$ but this integral is not convergent because the asymptotic behavior for very large values of $k$ is $\sim e^{ik(x-y)}$ and so the integrand does not decrease to zero.

Now the problem asks me to prove that there is only one bound state $\psi_0$ and to determine $\psi_0$ and his eigenvalue. The problem gives me also an hint

Use the result of the integral $\int_{-\infty}^{\infty} dk \langle x| \psi_k \rangle \langle y|\psi_k\rangle$ to guess the form of $\psi_0$.

But as I write above, that integral seems divergent!


If you add $\pm 1$ in the numerator you get a Dirac delta plus other pieces for which you can use residue theorem. Indeed: $\int_{-\infty}^{+\infty}dk \frac{k^2a^2}{k^2a^2+1}e^{ik(x-y)}= 2\pi\delta(x-y)-\int_{-\infty}^{+\infty}dk \frac{1}{k^2a^2+1}e^{ik(x-y)}$. Let $I$ be the last integral. The integrand function has two poles at $k=\pm i/a$ and it is $\mathcal{O}(k^{-2})$ for large $k$, so you can use residue theorem to evaluate $I$. Choosing the proper contour w.r.t the sign of $(x-y)$, you can verify that (if I'm not wrong) $$I=\pi a e^{-|x-y|/a} \, . $$

  • $\begingroup$ You're a new contributor so I'd suggest to you that while your one line answer is possibly all you wanted to say, we'd prefer you flesh it out - perhaps you could do some Mathjax explicitly showing /outlining the change you're suggesting. I suspect this might help get you more votes. In general we prefer more fleshed out answers. $\endgroup$ – StephenG Feb 29 '20 at 11:41
  • 2
    $\begingroup$ @StephenG but more importantly, we don't allow solutions to homework questions. $\endgroup$ – BioPhysicist Feb 29 '20 at 14:30
  • $\begingroup$ Thank you for your answer. The result of the complete integral is: $\delta (x-y) + \frac{1}{2a} e^{-\frac{x-y}{a}}(\textrm{tanh}(x/a)\textrm{tanh}(y/a) + \textrm{tanh}(y/a) - \textrm{tanh}(x/a) -1)$. If the basis was complete the result would have been only the Dirac delta. So I imagine I should write something like: $\int_{-\infty}^{\infty} dk \langle x| \psi_k \rangle \langle y|\psi_k\rangle + \langle x | \psi_0 \rangle \langle y| \psi_0 \rangle$ and, forcing this expression be equal to Dirac delta, find the form of $\psi_0$. Is this the right way? $\endgroup$ – dfgoe55 Feb 29 '20 at 15:49
  • $\begingroup$ I think so. Besides I guess that thinking to the properties of the potential (parity ecc..) could be useful to work out the shape of the bound state. (How many nodes do you expect it to have?). By the way, is that a Sissa problem? $\endgroup$ – Marco Feb 29 '20 at 15:58
  • $\begingroup$ Since it is a one dimensional problem with only a bound state, the wave function shouldn't have any node. I don't figure out how the previous result of the integral helps me to guess the form of $\psi_0$. Yes,it is a SISSA problem... $\endgroup$ – dfgoe55 Feb 29 '20 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.